What is the equation for the circle formed by the intersection of two spheres?

[songoku]

Homework Statement

In the space, consider the sphere S1 of radius 3 whose center is the point A (1, -1, 1) and the sphere S2 of radius 2 whose center is the point B(t, 1 - t, 1 + t).
a. Find the range of values of t in order the two spheres S1 and S2 have common points
b. Find the value of t for which B is closest to the point A
c. For the value of B obtained from (b), find the radius of circle formed as intersection of S1 and S2

Homework Equations

differentiation
equation of sphere: (x - a)² + (y - b)² + (z - c)² = r²

The Attempt at a Solution

I tried to equate the two equations of sphere and set discriminant ≥ 0, but I ended up having equation containing three variables x, y and z which can't be solved.

To be honest, the only equation I know about sphere is the one I wrote above...

 

[Dick]

It's not really all that much about the equation of the spheres. If the distance between A and B is greater then 5, then the spheres don't intersect. If it's less than 5 then they do. It's about the distance between A and B.

 

[songoku]

It's not really all that much about the equation of the spheres. If the distance between A and B is greater then 5, then the spheres don't intersect. If it's less than 5 then they do. It's about the distance between A and B.

Ah, why don't I think about it...

OK let's move on to find part (c). Actually I don't know which circle the question is referring to. I have image in my head that the intersection between two spheres are another 3 dimensional object, not a 2 dimensional object such as circle...

 

[HallsofIvy]

Go ahead and multiply out the two equations for the circles, then subtract one equation from the other. Since all "square" terms have coefficient 1, they will cancel giving a linear equation- the equation of the plane in which the two spheres intersect.

Then the circle of intersection is given by the simultaneous solution to the equation of either of the spheres and the equation of that plane.

 

[songoku]

Go ahead and multiply out the two equations for the circles, then subtract one equation from the other. Since all "square" terms have coefficient 1, they will cancel giving a linear equation- the equation of the plane in which the two spheres intersect.

Then the circle of intersection is given by the simultaneous solution to the equation of either of the spheres and the equation of that plane.

I found the equation of plane: 2y + 2z = 7

I can't find the simultaneous solution to the equation of sphere 1 and the plane. I substitute y = 7/2 - z to the equation of plane and end up having two variables, x and z? What should I do?

Thanks